Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z + 1}{-3z + 12} \times \dfrac{-z^2 + 2z + 63}{z^2 - 8z - 9} $
Explanation: First factor out any common factors. $q = \dfrac{z + 1}{-3(z - 4)} \times \dfrac{-(z^2 - 2z - 63)}{z^2 - 8z - 9} $ Then factor the quadratic expressions. $q = \dfrac {z + 1} {-3(z - 4)} \times \dfrac {-(z - 9)(z + 7)} {(z - 9)(z + 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(z + 1) \times -(z - 9)(z + 7) } {-3(z - 4) \times (z - 9)(z + 1) } $ $q = \dfrac {-(z - 9)(z + 7)(z + 1)} {-3(z - 9)(z + 1)(z - 4)} $ Notice that $(z - 9)$ and $(z + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-\cancel{(z - 9)}(z + 7)(z + 1)} {-3\cancel{(z - 9)}(z + 1)(z - 4)} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac {-\cancel{(z - 9)}(z + 7)\cancel{(z + 1)}} {-3\cancel{(z - 9)}\cancel{(z + 1)}(z - 4)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $q = \dfrac {-(z + 7)} {-3(z - 4)} $ $ q = \dfrac{z + 7}{3(z - 4)}; z \neq 9; z \neq -1 $